Question

Let A be an n # n matrix, b be a nonzero vector, and x0 be a solution vector of the system Ax D b. Show that x is a solution of the nonhomogeneous system Ax D b if and only if y D x!x0 is a solution of the homogeneous system Ay D 0.

Answer 1

Complete Question

Let A be an n x n matrix, b be a nonzero vector, and x_0 be a solution vector of the system Ax = b. Show that x is a solution of the non-homogeneous system Ax = b if and only if y = x - x_0 is a solution of the homogeneous system Ay = 0.

Answer:

Explanation:

From the question we are told that

A is an n × n matrix

b is a zero vector

`x_o` us the solution vector of
`Ax = b`

Which implies that

`Ax_o = b`

So first we show that

if
`x` is the solution matrix of
`Ax = b`

and
`y= x-x_o` is the solution of
`Ay = 0`

Then

`A(x-x_o) = 0`

=>
`Ax -Ax_o = 0`

=>
`b-b = 0`

Secondly to show that

if
`y= x-x_o` is the solution of
`Ay =0`

then x is the solution of the non-homogeneous system

`Ax = b`

Now we know that
`y = x-x_o` is the solution of
`Ay =0`

So

`Ay = 0`

=>
`A(x- x_o) = 0`

=>
`Ax - Ax_o = 0`

=>
`Ax - b = 0`

=>
`Ax = b`

Thus this has been proved